If a three-digit positive integer has its digits reversed, the

Question

If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

A. 3
B. 6
C. 7
D. 60
E. 70

M37-46

Bunuel · Accepted Answer

Official Solution:

If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

A.  3 
B.  6 
C.  7 
D.  60 
E.  70

Say the original three-digit number is  abc  and the reversed number is  cba .
 
Given:  abc-cba=297 ;
 
 (100a + 10b + c ) - (100c + 10b + c) = 297 ;
 
 a-c=3 .
 
 (a, c)  can be (9, 6), (8, 5), (7, 4), (6, 3), (5, 2), and (4, 1) (6 pairs). Notice that (6, 0) pair is not possible:  c  cannot be 0 because in this case  cba  would be a two-digit number, not a three-digit number as given in the stem.
 
Finally, don't forget about  b . It can take 10 values: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
 
So, there are a total of  6*10=60  numbers are possible.

Answer: D

sahilvijay · Answer

IMO D

abc -cba =297
100a+10b+c - (100c+10b+a) = 297
100(a-c) +c-a = 297
99(a-c) =297
a-c =3
b can take 0-9 => 10 values
but c can not be 0 since reverse number is also 3 digit number
c max = 6 because 6+3 =9 =a max
so c can take 1,2,3,4,5,6 => 6 values
Hence total numbers possible
6x10=60

D is the Answer

maray · Answer

100x+10y+z-100z+10y+x=297
99(x-z)=297
x-z = 3
there are 6 possible numbers for x. for example: 9y6-6y9=297 and so on till x=4
as for y. there are 10 possibilities: from 0 to 9
so 6*10=60 pairs possible. 
so answer is D

hiranmay · Answer

Ans: D N=100x+10y+z => (100x+10y+z) - (100x+10y+z) =297 => x-z =3 (1) x <=9, z <=6 <=> Total 6 pairs z=1,2,3,4,5 & 6 (2) y can be 0 to 9(Total 10) Total possibilities = 6*10=60

lor12345 · Answer

What about a?
b*c= 10 * 6= 60
so what about a? where have it gone?

sahilvijay · Answer

No need to consider a - above solution is self explanatory

Jerry1982 · Answer

Answer is D 
Possible combinations are 4b1, 5b2, 6b3,7b4,8b5, 9b6
B can be any dogit from 0-9 , hence 10 possiblities. and the above combination are only 6 , therefore answer is 6*10 = 60

gracie · Answer

xyz-zyx=297
297/99=3=x-z
6 possible x-z combinations: 9-6,8-5,7-4,6-3,5-2,4-1
10 possible y values: 0-9
6*10=60 possible pairs
D

Bunuel · Answer

Par of  GMAT CLUB'S New Year's Quantitative Challenge Set

Kinshook · Answer

Asked: If a three-digit positive integer has its digits reversed, the resulting three-digit positive integer is less than the original integer by 297. How many such pairs are possible?

Let the 3-digit positive integer be of the form xyz

(100x + 10y + z) - (100z + 10y + x) = 99(x-z) = 297
x -z = 3

Since x & z are not 0

(x,z) = {(4,1),(5,2),(6,3),(7,4),(8,5),(9,6)} = 6 cases
y can take 10 values

Total such numbers = 6*10 = 60

IMO D

exc4libur · Answer

ABC-CBA=297:100a+10b+c-100c-10b-a=297, 99(a-c)=297, (a-c)=3 
 (A,C)=(9,6;8,5;7,4;6,3;5,2;4,1)…(A,C)≠(3,0):C>0 
 (A,C)=6.pairs…B=(0,1,2,3,4,5,6,7,8,9)=10 
 Total.cases:6*10=60

Ans (D)

GmatPoint · Answer

﻿Considering the three-digit number to be ABC.
﻿The value of the three-digit number is given by 100*A+10*B+C.
﻿When the digits are reversed the number becomes CBA.
﻿The value becomes : 100*C + 10*B+A.
﻿Since the value decreased by 297 this can be represented as :
 ﻿\left(100A+10B+C\right)-\left(100C+10B+A\right)\ =\ 297 
﻿99(A-C) = 297.
﻿A-C = 3.
﻿The different possibilities include : (A = 9, C = 6), (A = 8, C = 5), (A = 7, C = 4), (A = 6, C = 3), (A = 5, C = 2), (A = 4, C = 1)
﻿A and C cannot take the value of 0 because then the numbers do not satisfy the three digit number condition.
﻿For the 6 case B can take any value from 0 to 9.
﻿A total of 10*6 = 60 cases.

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